Synopsys R&D Interview Question for Software Engineer / Developers


Country: India
Interview Type: Phone Interview




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what is the gaming pattern - round robin matches? or knock out? and to which level

- rajanikant.malviya September 26, 2012 | Flag Reply
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knock outs

- hari@hbiz.in September 26, 2012 | Flag
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A binary heap, with participants as leaves of the tree.
Split the two groups based on the ranking/seeding or whatever.
Eg: if n1,n2,..., n7,n8 are number according to ranking,
n1 is leaf in the first group and n2, the leaf in the second group
n3, n4 & so on...

The time complexity will be O(1) to find the winner and the runner.
Root of the heap is the winner.
The child element of the root, other than the winner is the runner up.

- MG September 26, 2012 | Flag Reply
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yes..i think u r correct

- Ramya September 26, 2012 | Flag
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No. Wrong. O(n-1) for winner and o(n+logn-2) for runner. If you don't understand, search for selection trees in geekforgeeks

- Hari@hbiz.in October 02, 2012 | Flag
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I guess MG is correct.. Selection tree will give the second ranked player not the runner up. Supposed the players are 1 ... 8 with 1 as lowest rank and 8 as highest rank.. suppose there is a match between 7 & 8 in the first round. and a match between 1 and 8 in the final.. and 8 wins.. MG's method will give runner up as 1 since 1 lost in final.. Selection tree will give answer as 7 since 7 is second highest ranked player...

- subhrob February 21, 2013 | Flag
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Linkedlist since any one can be removed at any time

- Ramya September 26, 2012 | Flag Reply
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It can be a binary heap

- Ramya September 26, 2012 | Flag Reply
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tournament trees or selection trees.

- KB September 28, 2012 | Flag Reply
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O(n) time to initialize the tree, O(1) to get the winner

- KB September 28, 2012 | Flag
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n1 n2 n3 n4 n5 n6 n7 n8

n1 n3 n5 n7

n1 n5


n1 ( winner)


can use max heap : who will match would be at root node so at the last winner to be at root
node .

- Amarkant Kumar October 01, 2012 | Flag Reply


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